PRACTICE PROBLEMS ON ARITHMETIC SEQUENCE AND SERIES

Problem 1 :

Which term of the AP

3/√7, 4/√7, 5/√7, ............ is 17/√7 ?

Solution

Problem 2 :

Divide 69 into three parts which are in A.P and are such that the product of the first two parts is 483.

Solution

Problem 3 :

Insert 4 arithmetic means between 4 and 324

Solution

Problem 4 :

If the pth term of an A.P is q and qth term is p the value of the rth term is 

(a)  p - q - r     (b) p + q - r    (c) p + q + r    (d)  None

Solution

Problem 5 :

The sum of a series in A.P is 72 the first term is 17 and the common difference -2, the number of terms is 

(a)  6     (b)  12     (c)  6 or 12     (d) None

Solution

Problem 6 :

The sum of all natural numbers between 500 and 1000, which are divisible by 13 is

(a)  28400     (b)  28405     (c)  28410     (d) None

Solution

Problem 7 :

If the 10th term of an A.P. is twice the 4th term, and the 23rd term is 'k' times the 8th term, then the value of 'k' is

(a) 2.5      (b) 3        (c) 3.5        (d) 4

Solution

Problem 8 :

The sum of the A.M. and G.M. of two positive numbers is equal to the difference between the numbers. The numbers are in the ratio.

(a) 1 : 3       (b) 1 : 6       (c) 9 : 1        (d) 1 : 12

Solution

Answer Key

1)  15th term of the arithmetic series is 17/√7.

2)  The required terms are 21, 23 and 25.

3)  68, 132, 196, 260

4)  t= q + p - r, option b.

5)  n = 6 and n = 12

6)  28405

7)  k = 2.5

8) 9 : 1

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